Question

Potassium chlorate is often used to generate oxygen gas in high school laboratory. If 183.7g of KClO3 is completely burnt catalytically, what volume of oxygen gas will be obtained at 39 celsius and 1200 torr pressure?

Answer 1

Question #68038, Physics / Molecular Physics

Potassium chlorate is often used to generate oxygen gas in high school laboratory. If 183.7g of KClO₃ is completely burnt catalytically, what volume of oxygen gas will be obtained at 39 celsius and 1200 torr pressure?

Solution:

Decomposition of Potassium chlorate:

$$2KClO_3 \rightarrow 2KCl + 3O_2$$

The mol number of O₂ is equal to:

$$2n(KClO_3) = 3n(O_2)$$ $$n(O_2) = \frac{2}{3}n(KClO_3)$$

The mol number of the Potassium chlorate can be calculated from the mass:

$$n(KClO_3) = \frac{m(KClO_3)}{M(KClO_3)} = \frac{183.7\,g}{122.55\,g/mol} = 1.5\,mol$$

Using Ideal gas law:

$$pV = nRT$$ $$p = 1200\,torr = 159986.8\,Pa$$ $$T = 39{}^\circ\mathrm{C} = 312.15\,K$$

From this equation volume of O₂:

$$V = \frac{nRT}{p} = \frac{2/3n(KClO_3) \cdot RT}{p}$$ $$V = \frac{\frac{2}{3} \cdot 1.5\,mol \cdot 8.314\,(cm^3 \cdot MPa/K \cdot mol) \cdot 312.15\,K}{0.16\,MPa} = 16220\,cm^3 = 16.22\,L$$

Answer: volume of oxygen is 16.22 L

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